Given that `E_(o_(2)//H_(2)O)^(Theta)= +1.23V`, `E_(S_(2)O_(8)^(2-)//SO_(4)^(2-))^(Theta)=2.05V`, `E_(Br_(2)//Br^(-))^(Theta)=+1.09V`, `E_(Au^(3+)//Au^(Theta))=+1.4V`
The strongest oxidizing agent is :

Options

A. `O_(2)`
B. `Br_(2)`
C. `S_(2)O_(8)^(2-)`
D. `Au^(3+)`

Answer

To determine the strongest oxidizing agent from the given standard reduction potentials, we will follow these steps:

### Step 1: List the given standard reduction potentials
We have the following standard reduction potentials:
1. \( E^\Theta_{O_2/H_2O} = +1.23 \, V \)
2. \( E^\Theta_{S_2O_8^{2-}/SO_4^{2-}} = +2.05 \, V \)
3. \( E^\Theta_{Br_2/Br^-} = +1.09 \, V \)
4. \( E^\Theta_{Au^{3+}/Au} = +1.40 \, V \)

### Step 2: Understand the relationship between reduction potential and oxidizing strength
The strength of an oxidizing agent is directly related to its standard reduction potential. The higher the standard reduction potential, the stronger the oxidizing agent. This is because a higher potential indicates a greater tendency to gain electrons (be reduced).

### Step 3: Compare the standard reduction potentials
Now we will compare the values:
- \( +1.23 \, V \) (for \( O_2/H_2O \))
- \( +2.05 \, V \) (for \( S_2O_8^{2-}/SO_4^{2-} \))
- \( +1.09 \, V \) (for \( Br_2/Br^- \))
- \( +1.40 \, V \) (for \( Au^{3+}/Au \))

### Step 4: Identify the highest potential
From the comparison:
- The highest standard reduction potential is \( +2.05 \, V \) for \( S_2O_8^{2-}/SO_4^{2-} \).

### Step 5: Conclude the strongest oxidizing agent
Since \( S_2O_8^{2-} \) has the highest standard reduction potential, it is the strongest oxidizing agent among the given options.

### Final Answer
The strongest oxidizing agent is \( S_2O_8^{2-} \).

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